A ball is thrown straight upward and returns to the thrower's hand after 2.00 s in the air. A second ball thrown at an angle of 40.0° with the horizontal reaches the same maximum height as the first ball.
(a) At what speed was the first ball thrown?
(b) At what speed was the second ball thrown?
Tr = Rise time, Tf = Fall time.
a. Tr + Tf = 2 s, Tf = Tr, Tr+Tr = 2, 2Tr = 2, Tr = 1 s.
V = Vo + g*Tr = 0, Vo = -g*Tr =
9.8*1 = 9.8 m/s.
b. Yo = 9.8 m/s(Part a), Vo*sin40
= 9.8, Vo = 9.8/sin40 = 15.25 m/s.
posted by Henry